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\begin{document}
\section{Special Relativity}
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Theoretical Tools:\\
1) Special Relativity\\
2) Quantum Mechanics.\\
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Coordinates: A point in space-time is specified by the coordinates of $ x^0 , x^1,x^2, and x^3, in which x^0 = ct, x^1 = x, x^2=y, x^3=z; $\\
The Covariant vector is defined as $ x_\mu = \eta_{\mu\nu} x^\nu$ 
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Introduce the 4 contravariant vector $x^\mu = (x^0,x^k) = (x^0,x^1,x^2,x^3)$.
where $\mu $= 0,1,2,3 and k= 1,2,3; \\
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\bsma
The covariant vector is defined as $ x_\mu = \eta_{\mu\nu}x_\nu = (t,\overline{x})$ , where $ \eta{\mu\nu} = diag(1,-1,-1,-1)$
\esma \\
extra identities				
\bea
 \eta_{\mu \nu} = \eta^{\mu \nu} \eta{_\mu}{^\nu} = \delta{_\mu}{^\nu} 
 x{_\mu}x^{\mu} = \eta_{\mu\nu}x^\mu x^\nu =  \eta_{00}x^0 x^0 + \eta_{11}x^1 x^1 +			 \eta_{22}x^2x^2 + \eta_{33}x^3x^3 \\  =  (x^0)^2 - (x^2)^2 -(x^2)^2 - (x^3)^2  =   t^2 - \overline{x}^2  
\eea
\textbf{$ {x_\mu}{x^\nu} = t^2 - \overline{x}^2 $}  - Einstein Summation Equation\\
\noindent The gradient operator is \\
$\partial{_\mu} = \frac{\partial}{\partial(x^\mu)} = \frac{1}{c}\frac{\partial}{\partial(t)},\overline{\nabla} $ \\
$\partial{^\nu} = \frac{\partial}{\partial(x_\nu)} = \frac{1}{c}\frac{\partial}{\partial(t)},\overline{\nabla} $ \\
The d'Alembertian operator is defined as \\
$   \partial{^\mu} \partial{_\mu} = \frac{1}{c^2} \frac{{\partial}^2}{\partial t{^2}}  $ \\
\noindent {\bf Lorentz transformation} \\
According to Special Relativity , the form of a theory describing nature, it has to be invariant under a transformation from one invariant frame to another.\
The observer in two different inertial frames S and S' and the coordinates of the event is $x_\nu$ wrt S and $x'_\nu$ wrt S'. \ The LT is a real linear transformation of the coordinates that conserves the norm of the intervals between all the points in space time. The LT $ \lambda{_\mu}{^\nu} $ which connects S and S' coordinates is given by $x'^{\mu} = \lambda^{\mu}_{\nu} x^{\nu} + a^{\mu}  $ where a is the translation of the space time axis. \\
\noindent We shall treat the translation $ a^{\mu} $ seperately and give the name of the LT as homogenous LT when $ a^{\mu} = 0$. The group formed by all the LT's including translations is called as inhomogenous Lorentz group or Poincare group. \\	\\
\textbf{Homogenous LT}
\noindent $x'^{\mu} = \lambda^{\mu}_{\nu} x^{\nu}$ 
\noindent  Boosts along the x direction 
\begin{center}
$ (x^0){'} = \frac{x^0 - \beta x'}{\sqrt{1-\beta^2}}$ where $\beta = v/c$ \\
$(x^1){'} = \frac{x^1 - \beta x^0}{\sqrt{1-\beta^2}}$ where $\beta = v/c$\\
$ (x^2){'} = x^2 $\\
$ (x^3){'} = x^3 $\\
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Rotation about the z axis \\
$(x^0){'} = x^0$\\
$(x^1){'} = x^1 cos\theta + x^2 sin\theta $ \\
$(x^2){'} = -x^1 sin\theta + x^2 cos\theta $ \\
$ (x^3){'} = x^3 $
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\noindent Problem 1. Calculate the LF for boost along x direction and rotation about the z axis. 
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Problem 2. Prove that $ \triangle$ is a lorentx scalar.\\
$ \partial{'}{_\mu} \partial{'}{^\mu} = \lambda{_\mu}{^\alpha} \partial{_\alpha} \lambda{^\mu}{_\beta} \partial{^\beta} = (\lambda{^T}){^\alpha}{_\mu} (\lambda){^\mu}{_\beta} \partial{_\alpha} \partial{^\beta} = (\lambda{^T} \lambda){^\alpha}{_\beta} \partial{_\alpha} \partial{^\beta} = \delta{^\alpha}{_\beta} \partial{_\alpha}{^\beta} = \partial{_\mu} \partial{^\mu}$\\
Hence we prove that $ \partial{_\mu} \partial{^\mu} $ is a lorentz scalar. \\
{\bf Classification }: \\
$ \eta{_\alpha}{_\beta} = \eta{_\mu}{_\nu} \lambda{^\mu}{_\alpha} \lambda{^\nu}{_\beta} $\\ 
now $ \eta{_\alpha}{_\beta}$ = diag(1,-1,-1,-1); say $ \alpha = 0, \beta = 0$ \\
% now we have to write some parts of classifications for lorentz factors 
\underline{Infinitesimal Lorentz transformations}\\
A simple approach to working with infinitisemal LT, we are to consider only proper orthogonal LT. \\
Consider the following ILT- 
$  \lambda{^\mu}{_\nu} = \delta{^\mu}{_\nu} + \omega{^\mu}{_\nu} $, where $\omega{^\mu}{_\nu} $ is the infinitisemal transfer parameter. \\
We know that $ x'{^\mu} x'{_\mu} = x{^\mu} x{_\mu} $ , which implies \\
$ \eta{_\alpha}{_\beta} = \eta{_\mu}{_\nu} \lambda{^\mu}{_\alpha} \lambda{^\nu}{_\beta} $\\ 
$ = \eta{_\mu}{_\nu} (\delta{^\mu}{_\alpha} + \omega{^\mu}{_\alpha}) (\delta{^\mu}{_\beta} + \omega{^\mu}{_\beta})$\\
$ =\eta{_\alpha}{_\beta} + \eta{_\mu}{_\nu} \delta{^\mu}{_\alpha} \omega{^\mu}{_\beta} + \eta{_\mu}{_\nu} \omega{^\mu}{_\alpha} \delta{^\mu}{_\beta}$\\
therefore $ \omega{_\alpha}{_\beta} + \omega{_\beta}{_\alpha} = 0$\\
hence $ \omega{_\alpha}{_\beta} = -\omega{_\beta}{_\alpha}$ so $\omega{_\beta}{_\alpha}$ is antisymmetric 
1. Components of $\omega{_\mu}{_\nu}$ \\
The number of independent components =  $\frac{n(n-1)}{2} = 6$ where n=4. A general LT thus has 6 parameters 3 veloctiy paramteres and 3 rotation parameters. \\
Rotation about the z axis . \\
$\lambda{^\mu}{_\nu}$ = \[ \left( \begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 &c{_\theta} & s{_\theta} & 0 \\
0 &-s{_\theta} & c{_\theta} & 0\\
0 & 0 & 0 & 1 \end{array} \right)\]
When $\theta$ is small we get\\
$\lambda{^\mu}{_\nu}$  = \[ \left( \begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & \theta & 0 \\
0 &-\theta & 1 & 0\\
0 & 0 & 0 & 1 \end{array} \right)\]\\
= $  \delta{^\mu}{_\nu} + \omega{^\mu}{_\nu} $ \\
= \[ \left( \begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1 \end{array} \right)\]   + 
\[ \left( \begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & 0 & \theta & 0 \\
0 & -\theta & 0 & 0\\
0 & 0 & 0 & 0 \end{array} \right)\] \\
Homework Construct LF for boost in x direction. \\
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\newpage 
% have to write a section on relativistic kinematics etc.
\section{Cross Section}
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Physical Quantities : Cross Section, Decay width. $\Rightarrow$ expressed in terms of invariant quantities. \\
Calcualte the 3 Mandelstam variables s,t and u defined as 
$ s=(p{_A}+p{_B})^2 = (p{_C}+p{_D})^2$ \\ 
$ t=(p{_A}-p{_C})^2 = (p{_D}+p{_B})^2 $\\ 
$ u=(p{_A}-p{_D})^2 = (p{_C}-p{_B})^2 $ \\ for a process $ A + B \rightarrow C + D $ . \\
Solution: By energy-momentum conservation we get,\\
$ p{_A} + p{_B} = p{_C} + p{_D} $  where $p_A$ .. are the 4 momenta. \\
Now, $ p_A ^\mu = (E_A,\vec{p_A}) ,p_B ^\mu = (E_B,\vec{p_B}) $\\
$ p{_A} \cdot p{_B} = E{_A}E{_B} -\vec{p_A} \cdot \vec{p_B} $. If the particles are on-shell we have $p_A^2 = m_A^2; p_B^2 = m_B^2; p_C^2 = m_C^2; p_D^2 = m_D^2$ Lorentz invariants\\
Hence $s+t+u= p_A^2 + p_B^2 + 2p{_A} \cdot p_B + p_A^2 + p_C^2 -2p_A \cdot p_C + p_A^2 + p_D^2- 2p_A \cdot p_D$\\ 
= $3m_A^2 + m_B^2 + m_C^2 + m_D^2 + 2p_A \cdot (p_B -p_C -p_D)$\\
= $ m_A^2 +m_B^2 + m_C^2+ m_D^2 $\\
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1. Consider the process $e^+ e^- \rightarrow \mu^+ \mu^-$ \\
We shall look at two types of experimental setups 
\begin{list}{}{}
\item 1. Fixed target 
\item 2. Colloding beam
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Lab Frame : $(E_1,\vec{p_1}) , (E_2,\vec{p_2})$ where $\vec{p_2} = 0 , so E_2 = m_2$ \\
COM Frame : $(E_1,\vec{p_1}) , (E_2,\vec{p_2})$ where $ \vec{p_1} = -\vec{p_2} and E_COM = E_1' + E_2' $\\
Let us assume $p_1 + p_2 = p $ and $p_1' + p_2' = p' $  \\
So, $p_\mu p^\mu = p_\mu' p{^\mu}{'} = ( $ \\

In the CM frame, $p_1 = (E,0,0,k) ; p_2=(E,0,0,-k) ;  p_3=(E,ksin\theta,0,kcos\theta) ; p_4 = (E,-ksin\theta,0,-kcos\theta)$, where $\vert\vec{k_i}\vert = \vert\vec{k_f}\vert = k $ \\
 $s= (p{_1}+p{_2})^2 = (E+E)^2 = 4E^2 $ \\
$ t= (p_1 - p_3)^2   = 2m^2- 2p_1\dot p_3  = 2m^2- 2[E^2 -0-0-k^2cos{\theta}] = -2k^2 +2k^2cos{\theta} $ 
$t = -2k^2(1-cos{\theta})$\\
and $u = (p_1 -p-4)^2 = 2m^2- 2p_1\dot p_4 = 2m^2 - 2(E^2 +k^2cos{\theta}$\\
 $u = -2k^2(1+cos{\theta})$
\section{Relativistic Wave Equation} 
Schrodinger eqn: \\
$ \bar{H} = \frac{\hat{p^2}}{2m} + V $  ,  E $ \rightarrow i \overline{h} \frac{\partial}{\partial t}$ , $ \overline{p} \rightarrow -i \overline{h} \overline{\nabla} $ \\
The non relativistic Schrodinger equation\\
$ i \overline{h} \frac{\partial \psi}{\partial t} = \bar{H} {\psi} = (-\frac{\bar{h}^2}{2m} \nabla^2 + V) \psi$ \\
\[ i \overline{h} \frac{\partial {\psi^\star}}{\partial t} = \bar{H} {\psi^\star} = (-\frac{\bar{h}^2}{2m} \nabla^2 + V) {\psi^\star} \] \\
multiplying the first equation by $\psi^\star$ and the second by $\psi$ and subtracting the two we get, \\
\[ i\bar{h}({\psi^\star} \frac{\partial \psi}{\partial t} + {\psi} \frac{\partial \psi^{\star}}{\partial t} ) = -\frac{\bar{h^2}}{2m} ({\psi^\star}\nabla^2 \psi) - (\nabla^2 {\psi^\star}\psi )\] \\
\[ i\bar{h} \frac{\partial({\psi^\star}\psi)}{\partial t} = -\frac{\bar{h}^2}{2m} [ {\psi^\star} \vec{\nabla} \cdot (\vec{\nabla} \psi) - \vec{\nabla} \cdot (\vec{\nabla} \psi) \psi] \] \\
Therefore \[ \frac{\partial(\psi^\star \psi)}{\partial t} = \vec{\nabla} \cdot [\frac{i \bar{h}}{2m} (\psi^\star \vec{\nabla} \psi)] \] \\
Hence \[ \frac{\partial \rho}{\partial t} + \frac{\vec{\nabla}}{\vec{j}}  = 0 \] where \[ \rho = \psi^\star \psi = |\psi|^2\] is the probability density. \\

%\bf{Discrete transformations}\\
Discrete transformations\\
When such a transformation is applied twice, one returns to the original state. It is described by a simple group of just two elements, the identity \textit{I} and the element \textit{g} where $g^2 = 1$\\
Examples are space inversion, particle conjugation. \\ 
Let U be a linear and unitary operator, then we have \[ U^\dagger = U^-1 \] \\
If U is to be an observable, it is also hermitian, hence \[ U^\dagger = U^-1 = U \] \\ After one application(in certain circumstances) the original state with utmost a phase factor us obtained, \[ p|i > = |pi> = {\eta_i} |i> ,  \eta  \] \\
When U = P, the parity operator, \\
\[ P^2|i> = {\eta_i}P |i> = {\eta_i}^2 |i> = |i>, \Rightarrow {\eta_i}^2 = 1, |{\eta_i}| = 1 \] \\
\[ {\eta_i}^2 = \eta^\star \eta = 1 , \Rightarrow \] \\
Lets study a few symmetries \\
Time reversal: Time reversal the name is a misnomer, actually the term reversal of motion explains it better. In a symmetry transformation the transition probabilty is to be preserved. The transformation can be unitary or anti-unitary. \\ For time reversal what one will need is a anti unitary operator.\\
Proof: Let the operator be U, which is unitary and it changes $ t \rightarrow -t $. If it is a symmetry, then \[ [U,H] = 0 \] i.e \[ UH\dagger{U} = H \] \\
The time dependant schrodinger equation is \[ i \overline{h} \frac{d \psi}{dt} = H \psi , \overline{h} =1 \] \\ \[ \rightarrow U i \frac{d \psi>}{dt} = UH \psi> , so  i(U \frac{d \dagger{U} U \psi>}{dt}) = (UH\dagger{U})(U \psi>) \] \\
hence \[ -i \frac{d U|\psi}{dt} = H(U|\psi>) \] \\ Define the operator T = UK, where K is an operator, for a more detailed analysis we refer to the discussion section 4.4 of sakurai, Modern quantum mechanics.
%So for discussing time reversal we set the weak requirement that \[ |< \tilda{\beta} | \tilda{alpha} > | = |<\beta | \alpha>| \] , this is not the only condition. the condition that \[ |<\tilda{\beta} | \tilda{alpha} > | = { | < \beta | \alpha > | }^\star  = |<\alpha | \beta>| \]
Include a separate section on discussion on time reversal symmetry. Needed to understand the T operator defined earlier. \\


Section{Relativistic single particle Klein gordan equation}

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